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Question : A band brake acts on the 3/4th of circumference of a drum of...

A band brake acts on the 3/4th of circumference of a drum of 450 mm diameter which is keyed to the shaft. The band brake provides a braking torque of 225 N-m. One end of the band is attached to a fulcrum pin of the lever and the other end to a pin 100 mm from the fulcrum. If the operating force is applied at 500 mm from the fulcrum and the coefficient of friction is 0.25, find the operating force when the drum rotates in the (a) anticlockwise direction, and (b) clockwise direction.

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Given : d = 450 mm or r = 225 mm = 0.225 m ; T_{ B }=225 N – m ; b = OB = 100 mm = 0.1 m ; l = 500 mm = 0.5 m ; \mu=0.25

Let     P = Operating force.

(a) Operating force when drum rotates in anticlockwise direction

The band brake is shown in Fig. 19.11. Since one end of the band is attached to the fulcrum at O, therefore the operating force P will act upward and when the drum rotates anticlockwise, as shown in Fig. 19.11 (b), the end of the band attached to O will be tight with tension T_{1} and the end of the band attached to B will be slack with tension T_{2}. First of all, let us find the tensions T_{1} \text { and } T_{2}.

We know that angle of wrap,

\theta=\frac{3}{4} \text { th of circumference }=\frac{3}{4} \times 360^{\circ}=270^{\circ}

 

=270 \times \pi / 180=4.713 rad

 

and       2.3 \log \left(\frac{T_{1}}{T_{1}}\right)=\mu . \theta=0.25 \times 4.713=1.178

 

\therefore      \log \left(\frac{T_{1}}{T_{2}}\right)=\frac{1.178}{2.3}=0.5123 \text { or } \frac{T_{1}}{T_{2}}=3.253      …(i)                                                         . . . (Taking antilog of 0.5123)

We know that braking torque  \left(T_{ B }\right),

225=\left(T_{1}-T_{2}\right) r=\left(T_{1}-T_{2}\right) 0.225

 

\therefore      T_{1}-T_{2}=225 / 0.225=1000 N       …(ii)

From equations (i) and (ii), we have

T_{1}=1444 N ; \text { and } \quad T_{2}=444 N

Now taking moments about the fulcrum O, we have

P \times l=T_{2} \cdot b \quad \text { or } \quad P \times 0.5=444 \times 0.1=44.4

 

\therefore P = 44.4 / 0.5 = 88.8 N

 

(b) Operating force when drum rotates in clockwise direction

When the drum rotates in clockwise direction, as shown in Fig.19.11 (a), then taking moments about the fulcrum O, we have

P \times l=T_{1} \cdot b \quad \text { or } \quad P \times 0.5=1444 \times 0.1=144.4

 

\therefore     P = 144.4 / 0.5 = 288.8 N

 

 

19.6
19.11